为什么一个char +另一个char =一个奇怪的数字 [英] why does a char + another char = a weird number

问题描述

以下是代码片段：

  public static void main（String [] arg）
 {
 char ca ='a'; 
 char cb ='b'; 
 System.out.println（ca + cb）; 
} 
  

输出为：

  195 
  

为什么会这样？我认为'a'+'b'将是ab， 12或 3 。

解决方案

+ code>是算术加法，不是字符串连接。您必须执行+ ca + cb ，或使用 String.valueOf 和 Character.toString 方法，以确保 + 的至少一个操作数是 String 将操作符设置为字符串连接。

JLS 15.18添加运算符

+ 运算符 String ，则操作是字符串连接。

否则， + 运算符的每个操作数的类型必须是可转换为原始数值类型的类型，否则会出现编译时错误。

至于为什么要获得195，这是因为在ASCII中，'a'= 97 和'b'= 98 和 97 + 98 = 195

---

这执行基本的 int 和 char 铸造。

  char ch ='a'; 
 int i =（int）ch; 
 System.out.println（i）; // prints97
 ch =（char）99; 
 System.out.println（ch）; // printsc
  

这忽略了字符编码方案的问题约...）。

---

注意，Josh Bloch指出，很不幸， + 对于字符串连接和整数加法重载（对于字符串连接重载+运算符可能是一个错误。 - Java Puzzlers 拼图11：最后笑）。

---

另请参阅

• 与空字符串连接做一个字符串转换真的那么糟吗？

Here's the code snippet:

public static void main (String[]arg) 
{
    char ca = 'a' ; 
    char cb = 'b' ; 
    System.out.println (ca + cb) ; 
}

The output is:

195

Why is this the case? I would think that 'a' + 'b' would be either "ab" , "12" , or 3.

Whats going on here?

解决方案

+ of two char is arithmetic addition, not string concatenation. You have to do something like "" + ca + cb, or use String.valueOf and Character.toString methods to ensure that at least one of the operands of + is a String for the operator to be string concatenation.

JLS 15.18 Additive Operators

If the type of either operand of a + operator is String, then the operation is string concatenation.

Otherwise, the type of each of the operands of the + operator must be a type that is convertible to a primitive numeric type, or a compile-time error occurs.

As to why you're getting 195, it's because in ASCII, 'a' = 97 and 'b' = 98, and 97 + 98 = 195.

---

This performs basic int and char casting.

 char ch = 'a';
 int i = (int) ch;   
 System.out.println(i);   // prints "97"
 ch = (char) 99;
 System.out.println(ch);  // prints "c"

This ignores the issue of character encoding schemes (which a beginner should not worry about... yet!).

---

As a note, Josh Bloch noted that it is rather unfortunate that + is overloaded for both string concatenation and integer addition ("It may have been a mistake to overload the + operator for string concatenation." -- Java Puzzlers, Puzzle 11: The Last Laugh). A lot of this kinds of confusion could've been easily avoided by having a different token for string concatenation.

---

See also

• Is concatenating with an empty string to do a string conversion really that bad?

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